SELECTED DISCUSSIONS TO exercises/problems
BELOW.
YOU ARE RESPONSIBLE FOR ALL PROBLEMS EVEN THE ONES WITHOUT DISCUSSIONS, WHICH ARE
DESIGNED TO HELP YOU DO RELATED EXERCISES. Each sublist is separated by
a semi-colon ";" and includes the name of a representative
example for those
exercises.

SELECTED DISCUSSIONS TO exercises/problems
BELOW. I've try to include interesting explanatory hyperlinks
throughout.
YOU ARE RESPONSIBLE FOR ALL PROBLEMS EVEN THE ONES WITHOUT DISCUSSIONS, WHICH ARE
DESIGNED TO HELP YOU DO RELATED EXERCISES. Each sublist is separated by
a semi-colon ";" and includes a representative example for
exercises. WE WILL REVIEW THEM IN-CLASS THE LAST WEEK BEFORE EXAMS.

(b) Force = ke2/r2 = m*a. Max a when r is
the smallest, the initial separation

17. E = slope of the line. Remember
to convert cm to meters.

28. (a) Surface is a plane and it
is at the center, between plates, parallel to the two
plates.
(b) Surface is a plane and it is (2/12)*6 cm to the right of neg.
plate.
(c) E = |gradient| = |total change in V/ total distance|

54. Note:
(1)The two capacitors in series near the a - terminal can be
combined to produce an net capacitance = C/2. You should be able
to prove this.
(2) Thus, closest to terminal a we have two capacitors
in parallel, C/2 and C. The net capacitance is C/2 + C = (3/2)*C.
(3) Thus we finally have two capacitors in series between
terminals a and b, C and (3/2)*C. The Cs (s means
series) is given by 1/Cs = 1/C +
1/[(3/2)*C] = 1/C + 2/(3C). Add these up to get the final
net-C.
Now, let us find the charges:

Go to step 3 . Before the final combo is computed, you have two C's in
series, C and (3/2)*C. Thus they have the same charge Q.
28 (V) = Q/C + Q/[(3/2)*C]. Solve for Q, the charge on the
bottom C in the figure.
To get the other charges: Note: Voltage across the (3/2)*C
is Q/[(3/2)*C], where Q is known. Thus , you can now get the
voltage = Q/[(3/2)*C] = 2Q/(3*C). This voltage equals the voltage
across the single C in that branch:
2Q/(3*C) = voltage across single C = Q'/C. Find Q' = charge on
single C .

FINALLY GET THE CHARGE ON THE ORIGINAL TWO CAPACITORS IN SERIES IN THE
UPPER BRANCH:
Since they are in series that have the SAME charge Q". The
voltage across these two C's is
2Q/(3*C) = Q"/C + Q"/C.
Find Q".

81.
(a) C = Area*epsilon/d
(b) Q = C*V, where V = V+ - V- , the
voltage difference between the two plates.
(c) E*d = V
(d) U = Q2/2C.
(e) If the plates are disconnected, you will still have the same charge
Q you had in the beginning, computed in part (b).
The force between the plates will be attractive so you would have to
pull the plates apart with a force, equal in magnitude to
the attractive one, to keep them from accelerating toward zero
separation.
C' = C = Area*epsilon/d', where d' is the
new separation. You can immediately find U = Q2/2C'.
Note since C' < C, you have larger energy in the capacitor.
That's because you did work to separate the plates, thus increasing the
potential energy of the plates. It's like lifting a ball above the
ground to increase its U = mgh. You can now find V' = Q/C', where
C' is the new capacitance. Note the voltage difference has increased to
V'. Finally, find E = V'/d' and you should note the magnitude of
the electric field E has not changed, also seen formula E =
Q/(A*epsilon). But you are increasing the energy in
the space between the plates because the volume of that region has
grown. Remember the formula
U = (1/2)*epsilon*E2*volume , where the energy density is
(1/2)*epsilon*E2 and the volume is
A*(separation distance). See equation 18.20.

68. We will review this more in class. The
main issue is this. The electric field inside the region between plates
where a dielectric resides is smaller than it would be in a vacuum.
That's because surface-charges of opposite sign and polarity
than plate's charges gather on the edges of the
dielectric next to plates. These "edge" charges cause a field in the
opposite direction that partially cancels out the vacuum field Eo. See
figure 18.33 and class notes:
E = Eo/k, where Eo is the vacuum field if there was no dielectric. C =
Q/V = Q/[(Eo/k)*d]. Since Eo = Q/(A*epsilon), we see C =
k*Co, where Co is the capacitance without a dielectric. C goes up
since V = (Eo/k)*d goes down.
(a) E = Eo < 30000 V/m. Use Eo = Q/(A*epsilon) to get Q.
(b) E = Eo/k < 30000 V/m. . Solve for limit on Eo and use Eo = Q'/(A*epsilon) to get
limit Q', a
larger charge value than Q..